Đáp án:
\(x > 2;x \ne 3\)
Giải thích các bước giải:
\(\begin{array}{l}
Xét:\dfrac{{x + 1}}{{{x^2} - 4x + 4}} \ge 0\\
DK:x - 2 \ne 0 \to x \ne 2\\
\dfrac{{x + 1}}{{{x^2} - 4x + 4}} = \dfrac{{x + 1}}{{{{\left( {x - 2} \right)}^2}}} \ge 0\\
\to x + 1 \ge 0\left( {do:{{\left( {x - 2} \right)}^2} > 0\forall x \ne 2} \right)\\
\to x \ge - 1;x \ne 2\\
Xét:\dfrac{{{{\left( {2 - x} \right)}^3}}}{{{{\left( {x - 3} \right)}^4}}} \le 0\left( {DK:x \ne 3} \right)\\
\to 2 - x \le 0\left( {do:{{\left( {x - 3} \right)}^4} > 0\forall x \ne 3} \right)\\
\to x \ge 2;x \ne 3\\
KL:x > 2;x \ne 3
\end{array}\)
